Pt_{(s)} | Fe^{+2}(10^{-2} M, aq), Fe^{+3}(1 | vedclass

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(A) The balanced cell reaction is: $10Fe^{+2} + 2MnO_4^- + 16H^{+} 	o 2Mn^{+2} + 10Fe^{+3} + 8H_2O$The number of electrons transferred $(n)$ is $10$.

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$-2.36$

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(A) The balanced cell reaction is: $10Fe^{+2} + 2MnO_4^- + 16H^{+} 	o 2Mn^{+2} + 10Fe^{+3} + 8H_2O$The number of electrons transferred $(n)$ is $10$.The reaction quotient $Q$ is given by: $Q = \frac{[Mn^{+2}]^2 [Fe^{+3}]^{10}}{[Fe^{+2}]^{10} [MnO_4^-]^2 [H^{+}]^{16}}$Given $[H^+] = 10^{-pH} = 10^{-1} \ M$. Substituting the values:$Q = \frac{(10^{-2})^2 (10^{-3})^{10}}{(10^{-2})^{10} (10^{-3})^2 (10^{-1})^{16}} = \frac{10^{-4} 	imes 10^{-30}}{10^{-20} 	imes 10^{-6} 	imes 10^{-16}} = \frac{10^{-34}}{10^{-42}} = 10^8$Using the Nernst equation: $E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log(Q)$$E_{cell} = -2.31 - \frac{0.0591}{10} \log(10^8)$$E_{cell} = -2.31 - 0.00591 	imes 8 = -2.31 - 0.04728 = -2.35728 \ V \approx -2.36 \ V$

$Pt_{(s)} | Fe^{+2}(10^{-2} \ M, aq), Fe^{+3}(10^{-3} \ M, aq) || MnO_4^-(10^{-3} \ M, aq), Mn^{+2}(10^{-2} \ M, aq) | Pt_{(s)}$
At $298 \ K$,$E^o_{cell}$ for the cell is $-2.31 \ V$. What will be the $E_{cell}$ at $pH = 1$? .......... $Volt$

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$Pt_{(s)} | Fe^{+2}(10^{-2} \ M, aq), Fe^{+3}(10^{-3} \ M, aq) || MnO_4^-(10^{-3} \ M, aq), Mn^{+2}(10^{-2} \ M, aq) | Pt_{(s)}$At $298 \ K$,$E^o_{cell}$ for the cell is $-2.31 \ V$. What will be the $E_{cell}$ at $pH = 1$? .......... $Volt$